void函数和float函数有什么区别？

Question

I am still learning but... I don't think these functions:

void CalculateGross(float hours, float payrate, float *gross) //3.4
float CalculateGross(float hours, float payrate)              //3.4

do the same thing. So which is better practice? I'm assuming void is better?

/*

    3.0 Payroll Application

      3.1 PrintReportHeadings(inout reportFile as file)
      3.2 InitializeAccumulators(out totreg, totovt,totpayrate,totgross,totfed,
                                     totstate,totssi,totdefr, totnet as real)
      3.3 InputEmployeeData(out lastname, firstname as string,
                            out hours, payrate, defr as real)
      3.4 CalculateGross(in hours, payrate as real, out gross as real)      
      3.5 computeTaxes(in g,d as real, out ft, st, ssit as real)
        3.5.1 cFed(in g,d as real, out fed as real)
        3.5.2 cState(in ft as real,out state as real)
        3.5.3 cSSI(in g,d as real, out ssi as real)    
      3.7 PrintSummaryReport( ......)
     3.7.1 printTotals( ....)
     3.7.2 printAverages( ....)

*/

#include <stdio.h>
#include "TAXRATES.h"

void InputEmployeeData(char *lastname,char *firstname, // 3.3
                       float *hours,float *payrate, float *defr);
void CalculateGross(float hours, float payrate, float * gross); //3.4
float CalculateGross(float hours,float payrate); //3.4
extern void computeTaxes(float g,float d,float * ft,float *st,float *ssit); //3.5

int main(void)
{
    char ln[15+1];
    char fn[10+1];
    float fed,state,ssi;
    float g,h,p,d,n;

    InputEmployeeData(&ln[0],&fn[0],&h,&p,&d); // call 3.3
    g =  CalculateGross(h,p); // call 3.4
    // vs
    //CalculateGross(40.00,25.00,&g); // alternate call 3.4 
    computeTaxes(g,d,ADDR(fed),ADDR(state),ADDR(ssi)); // call 3.5
    n = g-fed-state-ssi-d;
    printf("  Fed   =   %8.2f\n",fed);
    printf("  State =   %8.2f\n",state);
    printf("  SSI   =   %8.2f\n",ssi);
    printf("  Net   =   %8.2f\n",n);
    while(getchar() != '\n'); // flush(stdin)
    return 0;
}

void CalculateGross(float hours, float payrate, float * gross) //3.4
{
if (hours <= 40)
    *gross = hours * payrate;
  else
    *gross = 40* payrate + 1.5 * payrate * (hours-40);

}

float CalculateGross(float hours,float payrate) .. //3.4
{
  if (hours <= 40)
    return hours * payrate;
  else
    return 40* payrate + 1.5 * payrate * (hours-40);
}

void InputEmployeeData(char *lastname,char *firstname, // 3.3
                       float *hours,float *payrate, float *defr)
{
    printf(" Enter the name ==> ");
    scanf("%s%s",firstname,lastname);
    printf(" Enter the hours and payrate ==> ");
    scanf("%f%f",hours,payrate);
    printf("  Enter the deferred earning amount ==> ");
    scanf("%f",defr);
}

Answer 1

Functions should have input(s) and a single output. Void says do something. Pointers are great, but the intent is to calculate something. Use f(x) = y, not f(x, out y) when possible, of course.

Answer 2

void CalculateGross(float hours, float payrate, float * gross); //3.4
float CalculateGross(float hours,float payrate); //3.4

where:

void CalculateGross(float hours, float payrate, float * gross) //3.4
{
if (hours <= 40)
    *gross = hours * payrate;
else
    *gross = 40* payrate + 1.5 * payrate * (hours-40);

}

Here - it does NOT have any effect on your code. (you will get warnings.. but no impact) Why? The type before the function name defines the type of return for the function. What you get back if you do:

foo = CalculateGross (....);

Since CalculateGross does not return any value (it only operates on values internally), the return type has no impact anywhere else.

Which is correct? That is a different question. void is correct for all function where no value is returned. When you specify float , the compiler will check your function for return somevalue; . If not found it will generate a warning (or error).

So, if your function returns a value (or pointer), match the type to the type the function returns. If your function returns nothing, void is the proper return type.

Note: just because you declare a function void does not mean you cannot use return within the void function. You can use return; by itself to cause to function to return ( bail out ) at that point without returning a value.